Week 3 Homework Solutions PDF

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1. Inside a starship at rest on the earth, a ball rolls off the top of a horizontal table and lands a distance d 1 from the foot of the table. This starship now lands on the unexplored Planet X. The commander, Captain Curious, rolls the same ball off the same table with the same initial speed as on earth and finds that it lands a distance d 2 from the foot of the table. What is the acceleration due to gravity on Planet X? The only force is that of gravity, and it pulls down. Therefore, the horizontal speed is constant, and the distance the ball travels is directly proportional to the fall time. We can compute the time the ball spent in the air on earth (t1 ): 1 s = at2 + v0 t + s0 2 1 0 = − gt12 + h 2

(1) (2)

Solving this for t1 : t1 =

s

2h g

We can likewise compute the time the ball falls on Planet X, t2 : s 2h t2 = gX

(3)

(4)

where gX is the gravitational acceleration on Planet X. Lastly, we know that our horizontal velocity can be expressed in two ways: d1 t1 d2 v= t2

v=

(5) (6)

Setting them equal, d1 d2 = t1 t2 r r g gX d1 =d 2 2h 2h

(7) (8)

Square both sides, and solving for gX : d 21 g =d22gX

(9)

d2 gX = 21 g d2

(10)

It’s easy to verify that this answer is dimensionally correct, and makes physical sense. If the ball traveled significantly farther on Earth (i.e. d 1 ≫ d 2 ) then Planet X’s gravity must be much stronger.

2. The earth has a radius of 6380 km and turns around once on its axis in 24 h.

(a) What is the radial acceleration of an object at the earths equator? Give your answer in m/s2 and as a fraction of g . The radial acceleration for circular motion is a=

v2 r

The linear velocity v can be computed as rω, or v =rω v =6380km ∗

2π ≈ 464m/s 24h

Tossing this into our acceleration formula, we get a = 0.034m/s2 . As a fraction of g , this is three-tenths of a percent, or a = 0.0035g

(b) If arad at the equator is greater than g, objects will fly off the earths surface and into space. What would the period of the earths rotation have to be for this to occur? We can recast the acceleration formula, using v = 2πr/T : a=

4π 2 r v2 = T2 r

Since we want a = g, we can just solve for T : T = 2π

r

r g

This is dimensionally correct! Tossing in our numbers, T = 5067s or approximately 1.5 hours.

4. A compact disc (CD) stores music in a coded pattern of tiny pits 10−7 m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are Rinner and Router , respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed v.

(a) What is the angular speed of the CD when the innermost part of the track is scanned? The outermost part of the track? We know that v = rω v ω= r We also know that the linear velocity v is constant, so v Rinner v = Router

ωinner = ωouter

(b) The maximum playing time of a CD is ∆t. What would be the length of the track on such a maximum- duration CD if it were stretched out in a straight line? If we fed a straight track at velocity v through a reader for time ∆t, it would have to be at least v∆t long.

(c) What is the average angular acceleration of a maximum duration CD during its playing time? Take the direction of rotation of the disc to be positive. The average angular acceleration is hαi =

∆ω ∆t

We are told in the problem statement that the track “spirals outwards towards the rim of the disc”, so we start at Rinner . The change in angular velocity is therefore ∆ω = ωouter − ωinner so hαi = Since

1

<

1

1 ωouter − ωinner = ∆t ∆t



v Router

−

v Rinner



, our value for hαi is negative; our angular velocity is in fact decreasing. This is consistent Router Rinner with the behavior of a disc that starts on the inside.

5. You are prone to losing your keys. Like, you lose them all the time. To remedy this terrible affliction you have instituted a policy of putting your keys onto a particular shelf as soon as you get home. You just got home from a particularly grueling physics recitation and are slumped on the sofa, when you feel your keys in your pocket. You don’t want to get up so you decide to toss the keys onto the shelf which is a horizontal distance d away, and a vertical distance h above you. Thinking about it you realize you have a lot of options to get the keys to land on the shelf. You could throw the keys fast, at a steep angle, or more slowly, at a shallower angle. Find a relationship between the angle above horizontal θ and the speed v at which you launch your keys to ensure they land on the shelf. Plot a graph showing the range of possible angles and the corresponding launch speeds.

Let’s begin this problem by choosing a coordinate system and sketching a possible trajectory for the keys:

d Since we are to hit a particular target, the motion is constrained in both directions. That is: ∆𝑥 = 𝑑 ∆𝑦 = ℎ

The primary variable here is the initial velocity vector 𝑣𝑖 . Expressed in components, this is: 𝑣𝑖𝑥 = 𝑣 cos𝜃 𝑣𝑖𝑦 = 𝑣 sin𝜃

If we ignore air resistance, the keys are in free-fall and 𝑎𝑥 = 0 𝑎𝑦 = −𝑔

Consider the motion is x first:

∆𝑥 = 𝑣𝑖𝑥 ∆𝑡 +

1

2

𝑎𝑥 (∆𝑡)2

𝑑 = 𝑣cos𝜃∆𝑡 ∆𝑡 =

𝑑 𝑣cos𝜃

This expression relates the time that the keys spend in the air to the initial speed and angle (and also the horizontal displacement d). If the keys are to hit the target, they must undergo the required vertical displacement (∆𝑦 = ℎ) during that same time. Now consider the motion in y: 1 ∆𝑦 = 𝑣𝑖𝑦 ∆𝑡 + 𝑎𝑦 (∆𝑡)2 2 1 ℎ = 𝑣sin𝜃∆𝑡 − 𝑔(∆𝑡)2 2

Substituting out expression for the time that the keys are in the air, we have: ℎ = 𝑣sin𝜃

𝑑 1 𝑑 ) − 𝑔( 𝑣cos𝜃 2 𝑣cos𝜃

ℎ = 𝑑tan𝜃 −

2

𝑔𝑑2 2𝑣 2 cos2 𝜃

This expresses the conditions that must be met if the keys are to hit the target. If we rearrange we can see the relationship between initial speed v and launch angle 𝜃 𝑣=

𝑑 𝑔 √ cos 𝜃 2(𝑑tan𝜃 − ℎ)

We can plot this, perhaps using a graphics calculator with numerical values for d and h. The important thing here is the shape of the curve. Does it make sense and conform to our intuition?

v

vmin

θmin

θmax

θ

There are some interesting extremes here. As 𝜃 → 90° the required launch speed approaches infinity (out expression for v has a vertical asymptote at 𝜃 = 90°) This makes sense: one way of having the keys land on the shelf would be to launch them at a very steep angle, with a very high speed:

y

h

d

x

Our expression for v also has a vertical asymptote when 𝑑tan𝜃 = ℎ, so that the required launch ℎ speed approaches infinity as 𝜃 → tan−1 ( ) . 𝑑

This corresponds to launching the keys directly at the shelf:

y

h

d

x

Between these extremes the required launch speed is reduced, reaching a minimum half way between....